Download MOSFET Current Mirror Current Sources

MOSFET Current Mirror
Current Sources
Basic Concepts
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display.
Let us study the most basic Current Mirror circuit,
consisting of two matching MOSFET transistors
connected back-to-back, such that both have the same
Gate-to-Source voltage:
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Given:
(*) The two enhancement-type(增强型) NMOS transistors
have matching features, as follows: VTH,1 = VTH,2 , kn,1’ =
kn,2’, λ1 = λ2 =0
(*) “By structure” we have that VGS,1 = VGS,2
(*) Typically the supply VDD, resistor R and a desired
reference current Iref are all given.
(*) The ratio (W/L)1 is not necessarily equal to the ratio
(W/L)2
(*) “By structure” we have that VGD,1 = 0, and because the
transistor is an enhancement-type, this guarantees that
transistor 1 is always in Saturation Mode
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Need: Using the
transistors geometries
(W/L)1 and (W/L)2 as
design parameters, we
want to create a DC
current Io, as long as
transistor Q2 is in
Saturation Mode
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The Drain of transistor Q2 is connected to a load circuit, not
necessarily a resistor. The load circuit typically involves one
or more additional MOSFET transistors. Depending on the
load, transistor Q2 may be in any of three modes: Saturation,
Triode or Cutoff. Of course, only when it is in Saturation it
will work as originally planned (a DC current source)
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The current Io always goes away from the load
circuit and into Q2. Such a DC current source is
said to be a sink.（接收器）
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Design of a Current Mirror DC Sink
We shall look first at Q1:
I D ,1 = I ref
=
1
W
= k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2
2
L
VDD − VGS ,1
R
So, indeed if Iref is specified and VDD and R are given,
then by the right-hand term the needed voltage VGS,1 is
specified. Then, using the middle term, need to solve
for (W/L)1.
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Example 1:
Let VDD = 5V, VTH,1 = 1V, kn,1' = 20μA/V2 and R = 1KΩ.
What should be (W/L)1 needed for creating Iref = 1mA?
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Solution of Example 1:
I ref = 1mA =
⇒ I ref = 1mA =
⇒(
VDD − VGS ,1
R
=
5 − VGS ,1
1
⇒ VGS ,1 = 4V
W
W
1
1
k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2 = ⋅ 20 ⋅10 −3 ( )1 ( 4 − 1) 2 ⇒
L
L
2
2
W
)1 = 11.11
L
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Let us now focus our attention on the
"mirror" transistor Q2:
I D,2
1
W
2
= I o = k n , 2 ' ( ) 2 (VGS , 2 − VTH , 2 )
2
L
I D ,1 = I ref
1
W
= k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2
2
L
We now divide the two equations and use all the given
matching parameters of the two transistors.
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Io
I ref
W
) 2 (VGS , 2 − VTH , 2 ) 2
L
=
W
0.5k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) 2
L
0.5k n , 2 ' (
Obviously, many terms cancel out and we are left
with the following simple design formula:
Io
I ref
W
( )2
= L
W
( )1
L
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Example 2 (Follow-up to Example 1)
What should be (W/L)2
if we want Io = 7mA?
Solution
By Example 1 we have
that Iref = 1mA and
(W/L)1
=
11.11.
Therefore:
Io
I ref
W
W
( )2
( )2
7 mA
W
L
L
=
⇒
=
⇒ ( ) 2 = 77.77
W
1mA 11.11
L
( )1
L
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Question:
What do we do if we want to create a source DC current
source, rather than a sink ?
[A "source" is when the current goes from the current
source into the load circuit]
Answer:
If we build a current mirror current source using PMOS
transistors (rather than NMOS) then the output DC current
will be "sourced" and not "sunk".
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Question:
If we need to generate multiple different DC current
sources and sinks, what is the total number of resistors
needed for the design?
Answer:
Just one resistor for the entire circuit!
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Current Steering
The use of a negative DC supply –VSS does not change the
fact that, by-structure, both transistors, in every mirror pair,
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have the same VGS voltage.
Recalculation of Reference Current
I D ,1 = I ref
VDD − (−VSS + VGS ,1 )
W
1
2
= k n ,1 ' ( )1 (VGS ,1 − VTH ,1 ) =
L
R
2
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NMOS Current Mirror Sinks:
I2
I ref
W
)2
= L
W
( )1
L
(
I3
I ref
W
)3
= L
W
( )1
L
(
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Current Steering Mechanism:
The Drain current of the NMOS transistor Q3 comes from
the Drain of the PMOS transistor Q4.
I4 = I3
Can "steer" a current from NMOS current mirror to PMOS
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current mirror, or vice versa.
Current Steering Mechanism:
There is no need for the NMOS and PMOS to be matching:
Only all the NMOS transistors must match among
themselves, and the PMOS transistors must be mutually
matching.
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PMOS Current Mirror
For the PMOS transistors Q4 and Q5, the following
parameters must match:
VTH,4 = VTH,5 , kp,4’ = kp,5’, λ4 = λ5
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PMOS Current Mirror
W
( )5
I5
L
=
W
I4
( )4
L
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We can now relate the sourced
current I5 to the reference current Iref:
I5
I ref
W
W
)3
( )5
I
I I
= 3 ⋅ 4 ⋅ 5 = L ⋅1 ⋅ L
W
W
I ref I 3 I 4
( )1
( )4
L
L
(
I5
I ref
W
W
)3 ( )5
= L ⋅ L
W
W
( )1 ( ) 4
L
L
(
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1) We need only one resistor R, no matter how
large is the number of DC currents generated.
2) By using current steering we can create
sourcing currents from sinking currents or
vice versa.
3) The reference current can reside either in
a NMOS current mirror or in a PMOS
current mirror.
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Example (a follow-up to the previous example)
Given: VDD = 5V, VSS = 0
VTH,1 = VTH,2 = VTH,3 = 1V, kn,1' = kn,2'=kn,3'= 20μA/V2 and R =
1KΩ.
VTH,4 = VTH,5 = -1V, kp,4' = kp,5'= 30μA/V2
Need: A reference current of Iref = 1mA, one sink current of
7mA and one source current of 5mA.
Question: Using the diagram scheme just discussed, what
should be all (W/L) ratios of all five transistors?
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Solution
The beginning of the solution is identical to what has
been done in the previous examples.
Let's quote the results:
(W/L)1 = 11.11 and (W/L)2 = 77.77mA.
Current I2 is then the desired sink current
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Now:
I5
I ref
W
W
W
W
( )3 ( )5 ( )3 ( )5
5mA
=
= L ⋅ L = L ⋅ L
W
W
1mA
11.11 W
( )1 ( ) 4
( )4
L
L
L
There are infinitely many choices for the geometric
dimensions of transistors Q3, Q4 and Q5. For
instance, we may take (W/L)3 = 11.11, (W/L)4 = 10
and (W/L)5 = 50. Current I5 is then the required
source current.
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The channel-length modulation effect may be
responsible for errors in the operation of a Current
Mirror Current Source.
For instance, depending on the load of Q2 we may
get VDS,2 ≠ VDS,1. As a result the value of I2 will
slightly vary, depending on the load. It means that
the current source is not ideal - it has a finite output
resistance, equal to ro of the respective output
transistor.
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